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The question is in regards to the two lemniscatic elliptic functions, often called the 'sine lemniscate' and 'cosine lemniscate' functions. I have been trying to prove the following identity:

\begin{align} \frac{sl(x) sl'(y) + sl'(x) sl(y)}{1 + sl(x)^2 sl(y)^2} \end{align}

as equal to: \begin{align} \frac{sl(x) cl(y) + sl(y) cl(x)}{1 - sl(x)cl(y)sl(y)cl(x)} \end{align}

I have had no success. I've attempted using the following relations:

$$ (1+sl(x))(1+cl(x)) = 2 $$ $$ sl(x)^2 + cl(x)^2 + sl(x)^2cl(x)^2 = 1 $$ $$ sl(x) = \sqrt{\frac{1 - cl(x)^2}{1 + cl(x)^2}} $$ $$ sl(\frac{\varpi}{2} - x) = cl(x) $$ and while I've managed to mix things up nicely, I've not successfully gone from one identity to the other.

This has been vexing me for a bit. Can anyone please help?

Xander Henderson
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Simplex_
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1 Answers1

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The question is in regards to the two lemniscatic elliptic functions, often called the 'sine lemniscate' and 'cosine lemniscate' functions. I have been trying to prove the following identity: \begin{align} \frac{sl(x) sl'(y) + sl'(x) sl(y)}{1 + sl(x)^2 sl(y)^2} \end{align} as equal to: \begin{align} \frac{sl(x) cl(y) + sl(y) cl(x)}{1 - sl(x)cl(y)sl(y)cl(x)} \end{align} I have had no success.

Define, just for convenience, the product function

$$ \text{rl}(x) := \text{sl}(x) \text{cl}(x). \tag1 $$

A known property of the lemniscate elliptic functions is that

$$ \text{rl}(x)^2 = 1 - \text{sl}(x)^2 - \text{cl}(x)^2. \tag2 $$

Another known property is that

$$ \text{sl}'(x) = \text{cl}(x)(1 + \text{sl}(x)^2), \quad \text{cl}'(x) = -\text{sl}(x)(1 + \text{cl}(x)^2). \tag3 $$

Given variables $\,x\,$ and $\,y,\,$ define the abbreviated notations

$$ s_1 = \text{sl}(x),\qquad c_1 = \text{cl}(x),\qquad r_1 = \text{rl}(x),\\ d_1 = s_1' = c_1(1 + s_1^2),\quad q_1 = 1 - s_1^2 - c_1^2 - r_1^2, \\ s_2 = \text{sl}(y),\qquad c_2 = \text{cl}(y),\qquad r_2 = \text{rl}(y),\\ d_2 = s_2' = c_2(1 + s_2^2),\quad q_2 = 1 - s_2^2 - c_2^2 - r_2^2. \tag4 $$

Define the two further quantities

$$ s_3 := \frac{s_1 d_2 + d_1 s_2}{1 + s_1^2 s_2^2} , \qquad q_3 := \frac{s_1 c_2 + s_2 c_1}{1 - r_1 r_2}. \tag5 $$

Use elementary algebra to verify the algebraic identity

$$ (s_3-q_3)(1-r_1r_2)(1+s_1^2s_2^2)=(s_1s_2)(r_2q_1 + r_1q_2). \tag5 $$

Notice that $\,q_1 = q_2 = 0\,$ from equation $(2)$. This implies that the right side of equation $(5)$ is equal to $0$ and therefore $\,s_3 = q_3\,$ which was to be proved.

Compare my method with the one in the work by P. L. Robinson arXiv:1902.08614 titled "The Lemniscatic Functions"

We develop the theory of the lemniscatic functions sl and cl from their definition as solutions to an initial value problem.

on page $11$ also based on unique solution of initial value problems.

Somos
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    +1 I got confused thinking that $\text{cl} , z=\text{cn}(z, i) $ based on $\text{sl} , z=\text{sn} (z, i) $, but later checked that $\text{cl} , z=\text{cd} (z, i) $. – Paramanand Singh Jul 24 '23 at 01:47
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    @ParamanandSingh I highly recommend the very comprehensive Lemniscate elliptic functions Wikipedia article. – Somos Jul 24 '23 at 02:12
  • Thank you very much! I apologize for the tardiness in accepting, but I've not been online until recently. I very much look forward to working through this.

    And yes -- I second reviewing the excellent Wikipedia article. I've profited much from studying it and following-up on references.

     – Simplex_ Aug 06 '23 at 23:58
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    @Simplex_ Thanks for your question! It was good to renew my acquaintance with the Lemniscate functions. – Somos Aug 07 '23 at 02:12